Leonhard Euler, the great Swiss mathematician, discovered in the 18th century that for polyhedra in which any face is a hexagon or a pentagon and every vertex is of degree three there must be exactly twelve pentagonal faces. (The degree of a vertex are the number of edges that come together at that vertex.) This is referred to as Euler's Twelve Pentagon Theorem. For the proof of this theorem see Euler. There is a dual to this theorem to the effect that for a polyhedron with only triangular faces and all vertices are of degree five or six there must be exactly twelve vertices of degree five. For the proof of this theorem see Euler2. The material here deals with the analogues of the dual to Euler's Twelve Pentagon Theorem. The direct analogues of that theorem are dealt with at Euler Archimedean.

The Dual Theorems

For polyhedra with one type of face and vertices of possibly two different degrees there are cases in which analogs of the Euler case hold. Let n be the number of edges of the f polygon faces; i.e., the gonality of the faces. The number of faces is denoted as f and the number of edges of the polyhedron as e. The number of vertices of the two degrees are denoted as v_j and v_k. Euler's formula for polyhedra requires that

f - e + v_j + v_k = 2

A face-by-face count of the number edges yields a total of nf, but this has to be equal to 2e because each edge is counted exactly twice.

A vertex-by-vertex count of the edges yields jv_j+kv_k, but this has to be equal to 2e because each edge is counted twice, once at each end. This means

nf = 2e = jv_j+kv_k

From these equations it follows that

2(f-e) = −(n-2)f
and hence
−(n-2)f + 2v_j + 2v_k = 4

The two equations to be satisfied are then

−(n-2)f + 2v_j + 2v_k = 4
and
−nf + jv_j+kv_k = 0

The number of faces, f, can be eliminated by multiplying the first equation by n and the second equation by (n-2) and subtracting. The number of one type of vertex will be also eliminated, say v_k, to give a definite value to the number of the other type of vertex only if

2n = (n-2)k
or, equivalently
k = 2n/(n-2)

This the same form of the condition found previously. The degree of vertex, k, to be eliminated from the equations has to be 3, 4 or 6. The value of v_j then must satisfy the equation

[2n-(n-2)j]v_j = 4n
which reduces to
v_j = 4n/(2n-(n-2)j)

The value of n can be 3, in which case k=6. This is the dual Euler case when j=5. For v₅ the above formula gives the value 12. For the icosahedron there are 20 triangular faces with 12 vertices each of degree 5. Geodesic domes can be constructed from triangular faces and any number of degree 6 vertices and exactly 12 vertices of degree 5.

The value of n can be 4, in which case k=4.

The only polyhedron sastisfying this case is the cube in which j=3. For this case the formula gives 6 as the value of v₃. The cube is the only polyhedra sastisfying these conditions. In cube there are no vertices of degree 4 but 6 of degree 3.

For n=6, k must be 3. There appears to be no polyhedron that can be constructed entirely of hexagons. The volley ball and buckministerfullerenes involve polyhedra with many hexagons but there are 12 pentagons facilitating the arrangement.

(To be continued.)